An amusement park game, shown below, launches a marble toward a small cup. The m

ID: 1360777 • Letter: A

Question

An amusement park game, shown below, launches a marble toward a small cup. The

marble is placed directly on top of a spring-loaded wheel and held with a clamp. When

released, the wheel spins around clockwise at constant angular acceleration, opening the

clamp and releasing the marble after making 11/12 revolution. The top of the cup is level

with the centre of the wheel.

a) What angular acceleration is needed for the ball to land in the cup?

b) What angle does the ball’s trajectory make with the vertical at the instant it enters

the cup?

here first find where the marble is when it is released

theta = 2*pie * (12 / 12 - 11/12) = pie/6 rad = 30deg

x = 40 * sin(30deg) / 2 = 0.1 m

y = 20 * cos(30deg) = 0.1732 m

how tan does marble have to travel

delta x = 0.1 + 1 = 1.1 m

delta y = -0.1732 m

this is projectile motion it is launches at 30deg

what v needed to travel the distance

y = vy*t + 0.5 * a* t^2

-0.1732 = vy - 0.5 * g *t^2

x = x0 + vx + 0.5 * a * t^2

1.1 = 0 + vx + 0

t = 1.1 / vx

epress in terms of wf = t = 1.1 / wf * r * cos(theta) = t = 6.351 / wf

remove unknown t from the problem by substituting into y equation

0 = 0.1732 + wf * 0.2 * sin(30deg) * 6.351 / wf - 0.5 * 9.8 ( 6.351 /wf)^2

wf = 15.637 rad/sec

now we know wf - w0 and delta theta and we need to know

alpha = wf - w0 / delta t = wf / delta t

wf^2 = w0^2 + 2 * alpha * delta theta

wf^2 = 2 * alpha * delta theta

alpha = wf^2 / (2*delta theta)

alpha = 15.637^2 / ( 2 * 2 * 11 * pie / 12)

alpha = 21.2 rad/sec^2

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